Question

A particle moves towards east from a point A to a point B at the rate of $4km/h$ and then towards north from B to C at rate of $5km/h$. If $AB=12$ km and $BC=5$ km, then its average speed for its journey from A to C and resultant average velocity direct from A to C are respectively:

• A. $\frac{17}{4}$ km$/h$ and $\frac{13}{4}$ km$/h$
• B. $\frac{13}{4}$ km$/h$and $\frac{17}{4}$ km$/h$
• C. $\frac{17}{9}$ km$/h$ and $\frac{13}{9}$ km$/h$
• D. $\frac{13}{9}$ km$/h$ and $\frac{17}{9}$ km$/h$

Answer 1

The correct option is A $\frac{17}{4}$ km$/h$ and $\frac{13}{4}$ km$/h$

Given $AB=12km\left(s_1\right)$ and $BC=5km\left(s_2\right)$
Speed from $A$ to $B$ $=4km/h$
Time taken $t_1=\frac{12}{4}=3hr$
Speed from $B$ to $C$ $=5km/h$
Time taken to complete distance from $B$ to $C$, $t_2=\frac{5}{5}=1hr$

Average speed $=\frac{s_1+s_2}{t_1+t_2}=\frac{12+5}{3+1}=\frac{17}{4}km/h$

$AC=\sqrt{{\left(AB\right)}^2+{\left(BC\right)}^2}=\sqrt{144+25}=\sqrt{169}=13km$

Average velocity $=\frac{13}{4}km/h$